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Maximum length of string formed by concatenation having even frequency of each character

  • Last Updated : 13 Oct, 2021

Given N strings, print the maximum length of the string and the string formed by concatenating any of the N strings, such that every letter in the string occurs even number of times 

Example: 

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Input: N = 5, str = [“ABAB”, “ABF”, “CDA”, “AD”, “CCC”]
Output: ABABCDAADCCC 12
Explanation: The string formed by concatenation is ABABCDAADCCC. Each letter in the string occurs even number of times



Input: N = 3, str = [“AB”, “BC”, “CA”]
Output: ABBCCA 6
Explanation: The string formed by concatenation of all 3 strings is ABBCCA

 

Approach: The given problem can be solved using recursion and backtracking. The idea is to either include the string or exclude the string at every iteration. After including a string, the frequency of all the characters in the concatenated string is calculated. If frequency of all the characters is even we update the maximum length max. Below steps can be followed to solve the problem:

  • Initialize variable max to 0 for calculating maximum length of concatenated string having even frequency of all characters
  • Initialize string ans1 to store the concatenated string of maximum length with all character having even frequency
  • The base case of the recursive call is to return, if index becomes equal to the size of the input string list
  • At every recursive call we perform the following operation:
    • Include the string and check if the frequency of characters is even for the concatenated string
      • If the frequency is even, update max and ans1
      • Increment the index and make the next recursive call
    • Exclude the string, increment the index and make the next recursive call

Below is the implementation of the above approach:

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
int maxi = 0;
string ans1 = "";
 
// Function to check the string
void calculate(string ans)
{
 
    int dp[26] = { 0 };
    for (int i = 0; i < ans.length(); ++i) {
 
        // Count the frequency
        // of the string
        dp[ans[i] - 'A']++;
    }
 
    // Check the frequency of the string
    for (int i = 0; i < 26; ++i) {
        if (dp[i] % 2 == 1) {
            return;
        }
    }
    if (maxi < ans.length()) {
 
        // Store the length
        // of the new String
        maxi = ans.length();
        ans1 = ans;
    }
}
 
// Function to find the longest
// concatenated string having
// every character of even frequency
void longestString(vector<string> arr, int index,
                   string str)
{
 
    // Checking the string
    if (index == arr.size()) {
        return;
    }
 
    // Dont Include the string
    longestString(arr, index + 1, str);
 
    // Include the string
    str += arr[index];
 
    calculate(str);
    longestString(arr, index + 1, str);
}
 
// Driver code
int main()
{
    vector<string> A
        = { "ABAB", "ABF", "CDA", "AD", "CCC" };
   
    // Call the function
    longestString(A, 0, "");
 
    // Print the answer
    cout << ans1 << " " << ans1.length();
 
    return 0;
}
 
// This code is contributed by Potta Lokesh

Java




// Java Implementation of the above approach
 
import java.io.*;
import java.util.*;
 
public class index {
    static int max = 0;
    static String ans1 = "";
 
    // Function to check the string
    static void calculate(String ans)
    {
 
        int dp[] = new int[26];
        for (int i = 0; i < ans.length(); ++i) {
 
            // Count the frequency
            // of the string
            dp[ans.charAt(i) - 'A']++;
        }
 
        // Check the frequency of the string
        for (int i = 0; i < dp.length; ++i) {
            if (dp[i] % 2 == 1) {
                return;
            }
        }
        if (max < ans.length()) {
 
            // Store the length
            // of the new String
            max = ans.length();
            ans1 = ans;
        }
    }
 
    // Function to find the longest
    // concatenated string having
    // every character of even frequency
    static void longestString(
        List<String> arr, int index, String str)
    {
 
        // Checking the string
        if (index == arr.size()) {
            return;
        }
 
        // Dont Include the string
        longestString(arr, index + 1, str);
 
        // Include the string
        str += arr.get(index);
 
        calculate(str);
        longestString(arr, index + 1, str);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        ArrayList<String> A = new ArrayList<>();
        A.add("ABAB");
        A.add("ABF");
        A.add("CDA");
        A.add("AD");
        A.add("CCC");
 
        // Call the function
        longestString(A, 0, "");
 
        // Print the answer
        System.out.println(ans1 + " "
                           + ans1.length());
    }
}
Output
ABABCDAADCCC 12

 
 

Time Complexity: O(M*N* (2^N)), where N is the number of strings and M is the length of the longest string
Auxiliary Space: O(N)

Another Approach: The above approach can be further optimized by precomputing the frequency of characters for every string and updating the frequency array after concatenation of each string.

 




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